A student fires a bow and arrow in gym class and all his arrows land close to eachother but not on the bullseye. this student could be said to be:
precise but not accurate
Answer:
0.027 mole of NaOH.
Explanation:
We'll begin by obtaining the number of mole H2SeO4 in 45mL of 0.30M H2SeO4
This is illustrated below:
Molarity of H2SeO4 = 0.3M
Volume of solution = 45mL = 45/1000 = 0.045L
Mole of H2SeO4 =...?
Mole = Molarity x Volume
Mole of H2SeO4 = 0.3 x 0.045
Mole of H2SeO4 = 0.0135 mole
Next, the balanced equation for the reaction. This is given below:
H2SeO4 + 2NaOH –> Na2SeO4 + 2H2O
From the balanced equation above,
1 mole of H2SeO4 required 2 moles of NaOH.
Therefore, 0.0135 mole of H2SeO4 will require = 0.0135 x 2 = 0.027 mole of NaOH.
Therefore, 0.027 mole of NaOH is needed for the reaction.
Answer:
Making models
Explanation:
Classifying is the grouping together of items that are alike in some way. Answer: Making models involves creating rep- resentations of complex objects or processes.
Answer:
1.30 g
Explanation:
La reacción que toma lugar es:
Primero <u>convertimos 3.28 g de MgO en moles</u>, usando su <em>masa molar</em>:
- 3.28 g ÷ 40.3 g/mol = 0.081 mol MgO
Después <u>convertimos 0.081 moles de MgO en moles de O₂</u>, usando los <em>coeficientes estequiométricos</em>:
- 0.081 mol MgO *
= 0.0407 mol O₂
Finalmente<u> convertimos 0.0407 moles de O₂ en gramos</u>:
- 0.0407 mol O₂ * 32 g/mol = 1.30 g
Answer: 136 g of 
must be reacted to inflate an air bag to 70.6 L at STP.
Explanation:
Using ideal gas equation:
P= pressure of nitrogen gas = 1 atm (at STP)
V =volume of nitrogen gas = 70.6 L
n = number of moles of nitrogen gas = 1 atm (at STP)
R = gas constant = 0.0821 Latm/Kmol
T = temperature of nitrogen gas = 273 K (at STP)
For the balanced chemical reaction:
3 moles of nitrogen are produced by = 2 moles of
3.14 moles of nitrogen are produced by =
moles of
Mass of = Moles × Molar mass = 2.09 mole × 65 g/mol = 136 g
